Re: Loop-gain calculation

So - the question remains: What is "B" in your circuit?

Does the shown circuit represent the complete feedback amplifier? If so, the topology is different from the basic "loop gain A, feedback factor B" topology by an additional forward gain factor C. You get a closed loop gain expression like

Re: Loop-gain calculation

Consider a setup without additional external feedback, e.g. Vin connected to a voltage source. Then the internal feedback through R2 (feedback factor R1/(R1 +R2) ) is the only feedback in this circuit. It's not "marginal" but - among other parameters - determining the circuit gain. A loop gain can be calculated.

Re: Loop-gain calculation

Originally Posted by Easy peasy

is there is some marginal internal feedback? - nope. Loop gain refers to gain around the complete loop, there is no ckt shown to complete the loop.
what you are seeking is the forward transfer function of input to output with specified load.

The two transistors can be used as a linear amplifier.
The same applies to an opamp.
However, the problem in both cases is that - without negative feedback - there will be no usable DC bias point in the middle of the quasi-linear region.
(I know - the reason for the missing bias point may be not the same...but this is not of any importance with respect to the subject under discussion).

Therefore, we use two resistors (forming a simple voltage divider) for feeding back a part of the developped output voltage to the inverting input.
However, it is to be mentioned that - in the circuit under discussion - there must be 100% feedback for DC (CMOS inverter with Dcin=DCout).
Therefore, an additional capacitor at the input will be necessary. This results in a feedback factor k=1 for DC (no DC current through R2) and k=R1/(R1+R2) for AC.
For calculating the gain of the complete circuit, we are using the classical formula for the closed-loop gain of an amplifier with feedback (given in post#5): G=C*A/(1+A*B).

Re: Loop-gain calculation

a loop requires connection to the actual input

Why particularly the "actual" input? A loop is created by feeding back a signal from amplifier output to input. In the given circuit, the drain node can be recognized as output and the gate node as input.

Re: Loop-gain calculation

What is an "actual input"? What is your definition? The CMOS inverter has only one single input node - the common node for both gates.
Is there any doubt that R2 is connected between output and input?
As an example, consider the non-inverting opamp configuration.
The feedback path is connected to the inverting input - and the "actual" input (do you mean: signal input?) is the non-inv. opamp terminal.
The circuit under discussion shows the classical negative feedback scheme.
(by the way - "degeneration" is nothing else than negative feedback; see the "degeneration resistor" RE in the emitter path of a common emitter gain stage).

Re: Loop-gain calculation

I had a professor in undergrad who absolutely insisted that simple circuits like this be analyzed in terms of shunt/series feedback systems. What fun we had, spending half an hour analyzing calculating the gain of an inverting opamp amplifier, which could be solved in seconds using KVL and KCL.